# I'm so long on this one...

• Sep 25th 2007, 10:34 AM
circuscircus
I'm so long on this one...
If f is one-to-one, twice differentiable fucntion with inverse function g, show that

$\displaystyle g''(x) = -\frac{f''(g(x))}{[f'(g(x))]^3}$

What is this asking me to do?

Also deduce that if f is increasing and concave upward, hten its inverse function is concave downwards. (But I need to understand the top part before worrying about this)

I'm so lost...
• Sep 25th 2007, 10:53 AM
Jhevon
Quote:

Originally Posted by circuscircus
If f is one-to-one, twice differentiable fucntion with inverse function g, show that

$\displaystyle g''(x) = -\frac{f''(g(x))}{[f'(g(x))]^3}$

What is this asking me to do?

Also deduce that if f is increasing and concave upward, hten its inverse function is concave downwards. (But I need to understand the top part before worrying about this)

I'm so lost...

no, it is asking you to derive the formula given. i'll do the hard part.

Consider the function $\displaystyle f \left( g(x) \right)$

since $\displaystyle g(x) = f^{-1}(x)$, we have that:

$\displaystyle f \left( g(x) \right) = x$

differentiating both sides implicitly with respect to $\displaystyle x$, we get:

$\displaystyle f' \left( g(x) \right)g'(x) = 1$ .....................................note, we used the chain rule on the LHS

differentiating implicitly again, we obtain:

$\displaystyle f'' \left( g(x) \right)g'(x)g'(x) + f' \left( g(x) \right)g''(x) = 0$ ......note, we used the product rule (and chain rule) on the LHS

solving for $\displaystyle g''(x)$, we get:

$\displaystyle g''(x) = - \frac {f'' \left( g(x) \right) \left[ g'(x) \right]^2}{f' \left( g(x) \right)}$

Now, all you need to do, is prove that $\displaystyle \left[ g'(x) \right]^2 = \frac 1{ \left[ f' \left( g(x) \right) \right]^2}$ and you will have the desired result
• Sep 25th 2007, 11:05 AM
circuscircus
Ohh

Since g'(x) = 1/f'(g(x))

then square that, ok

Ok thanks! For the second part, I know that since it inverse, then it reflects on the y=x so if it is like a U shape, then reflecting it will make it concave down but how do I prove that using math on paper?
• Sep 25th 2007, 11:14 AM
Jhevon
Quote:

Originally Posted by circuscircus
Ohh

Since g'(x) = 1/f'(g(x))

then square that, ok

Ok thanks! For the second part, I know that since it inverse, then it reflects on the y=x so if it is like a U shape, then reflecting it will make it concave down but how do I prove that using math on paper?

Hint:

If $\displaystyle f$ is concave up, what does that say about $\displaystyle f''$?

if $\displaystyle g$ is concave down, what does that say about $\displaystyle g''$?
• Sep 26th 2007, 07:26 PM
circuscircus
f'(x) > 0

f''(x) > 0

g''(x) < 0

I can't really deduce anything from this...what else am i missing?
• Sep 26th 2007, 07:34 PM
Jhevon
Quote:

Originally Posted by circuscircus
f'(x) > 0

f''(x) > 0

these are what you are given.

Quote:

g''(x) < 0
this is your conclusion based on the formula you have. (g''(x) = - (positive quantity)/(positive quantity) which is obviously negative) mission complete.